Problem: Perform the row operation, $-2R_1\rightarrow R_1$, on the following matrix. $\left[\begin{array} {ccc} -7 & -9 & 4 & 6 \\ 0 & -6 & -4 & 0 \\ 2 & -6 & 2 & 1 \end{array} \right] $
Answer: Background There are three basic row operations that can be performed on matrices. $R_i \leftrightarrow R_j$. This symbol tells us to interchange rows $i$ and $j$. $cR_i \rightarrow R_i$. This symbol tells us to multiply a row $i$ by a constant $c$. $R_i + cR_j \rightarrow R_i$. This symbol tells us to add $c$ times row $j$ to row $i$. Finding the new row to be used For the given matrix, $R_1$ is given below. $R_1=\left[\begin{array} {ccc} -7 & -9 & 4 & 6 \end{array} \right]$ We are asked to perform the row operation, $-2R_1\rightarrow R_1$. Therefore, we must multiply $R_1$ by $-2$. $\begin{aligned}-2R_1 &= -2\left[\begin{array} {ccc} -7 & -9 & 4 & 6 \end{array} \right] \\\\&=\left[\begin{array} {ccc} 14 & 18 & -8 & -12 \end{array} \right]\end{aligned}$ Substituting the row Now, we must substitute row $R_1$ with $-2R_1$. $\left[\begin{array} {ccc} {-7} & {-9} & {4} & {6} \\ 0 & -6 & -4 & 0 \\ 2 & -6 & 2 & 1 \end{array} \right] \xrightarrow{-2R_1\rightarrow R_1} \left[\begin{array} {ccc} {14} & {18} & {-8} & {-12} \\ 0 & -6 & -4 & 0 \\ 2 & -6 & 2 & 1 \end{array} \right] $ Summary Our resultant matrix is the following. $ \left[\begin{array} {ccc} 14 & 18 & -8 & -12 \\ 0 & -6 & -4 & 0 \\ 2 & -6 & 2 & 1 \end{array} \right] $